chapter 18: thermodynamics renee y. becker valencia community college
TRANSCRIPT
Thermodynamics
Thermodynamics: The study of interconversion of heat and other forms of energy
Spontaneous process: Process that proceeds on its own without external influence (-G)
Non-spontaneous: Needs continuous external influence to take place (+G)
Thermodynamics
Entropy (S): Disorder, molecular randomness
S = Sfinal - Sinitial
When disorder increases +SWhen disorder decreases -S
Enthalpy (H): Heat flowIn to the system +HOut of the system -H
Gibbs Free-Energy: Measure of spontaneityG = H-TS
Example 1 Predict the sign of S in the system for each of the
following
a) H2O(g) H2O(l)
b) I2(g) 2I-(g)
c) CaCO3(s) CaO(s) + CO2(g)
d) Ag+(aq) + Br-
(aq) AgBr(s)
Which of the following reactions has an increase in entropy?
1. H2O(g) H2O(l)
2. H2O(l) H2O(g)
3. H2O(g) H2O(s)
Entropy and temperature 3rd Law of Thermodynamics
a) The entropy of a perfectly ordered crystalline substance at 0 K is zero
b) As the temperature increases, the KE increases, Molecular motion increases, entropy increases
Standard Molar Entropies and Standard Entropies of Reaction
Standard Molar Entropy, SThe entropy of one mole of the pure substance at 1 atm pressure and a specific temperature usually 25C
Standard Entropy of Reaction, SEntropy change for a chemical reaction
S = Sproducts - Sreactants
Based on 1 mole of substance so you have to multiply S by the number of moles present
Standard Entropy of Reaction, S
aA + bB cC + dD
S = [c S(C) + d S(D)] – [a S(A) + b S(B)]
Units: coefficients are moles
S = J/K mol
S = J/K
Example 2
Calculate the standard entropy of reaction at 25C for the decomposition of calcium carbonate
CaCO3(s) CaO(s) + CO2(g)
Substance S (J/K mol)
CaCO3(s) 92.9
CaO(s) 39.7
CO2(g) 213.6
Entropy and the Second Law of Thermodynamics
1st Law of Thermodynamics
In any process, spontaneous or nonspontaneous, the total energy of a system and its surroundings is constant
2nd Law of Thermodynamics
In any spontaneous process, the total entropy of a system and its surroundings always increases
Entropy and the Second Law of Thermodynamics
Stotal = Ssystem + Ssurroundings
if S > 0 spontaneous
if S< 0 non spontaneous
if S = 0 equilibrium
Ssurr = -H / T
Entropy and the Second Law of Thermodynamics
a) Exothermic reaction: H<0, because the surroundings gain heat (entropy increases), heat is lost from the
system
b) Endothermic reaction: H>0, surroundings lose heat
(entropy decreases), and system gains the heat
Which of the following reactions is endothermic?
1. N2O4(g) 2 NO2(g) H = +57.1 kJ
2. 2 NO2(g) N2O4(g) H = -57.1 kJ
Free-Energy
Free energy, G
G = H – TS
G = H - TS
if G < 0 spontaneous Only direct relationship that if G > 0 nonspontaneous always holds up!
if G = 0 equilibrium
Example 3 Consider the decomposition of gaseous N2O4
N2O4(g) 2 NO2(g) H = +57.1 kJ
S = +175.8 J/K
a) Is this reaction spontaneous under standard-state conditions at 25C?
b) Estimate the temperature at which the reaction becomes spontaneous
Standard Free-Energy Changes for Reactions 1. Standard State Conditions: Solids, liquids, and
gases in pure form at 1 atm pressure, Solutes at 1M concentration, specified temperature, usually 25 celsius
2. Standard Free Energy Change, G: The change in free energy that occurs when reactants in their SS are converted to products in their SS
3. G = H - TS G = H - TS
Example 4
Consider the thermal decomposition of calcium carbonate
CaCO3(s) CaO(s) + CO2(g) H = 178.3 kJ
S = 160.4 J/K
a) Calculate the standard free energy change for this reaction at 25Cb) Will a mixture of solid CaCO3, CaO, and gaseous CO2
at 1 atm pressure react spontaneously at 25C?
c) Assuming that H and S are independent of temperature, estimate the temperature at which the reaction becomes spontaneous
Standard Free Energies of Formation 1. Standard Free Energy of Formation, Gf
The free energy change for formation of one mole of the substance in its standard state from the most stable form of its constituent elements in their standard states
2. Gf measures the substances thermodynamic stability
with respect to its constituent elements
3. -Gf are stable and do not decompose to their
constituent elements under standard state conditions
Standard Free Energies of Formation
4. +Gf are thermodynamically unstable with respect to their
constituents elements
a) There is no point in trying to synthesize a substance that has a +Gf because it would degrade into it’s constituents
b) You would need to synthesize it at different temperatures and or pressures or start with different starting materials that has a reaction with a -Gf
5. G = Gf(products) - Gf(reactants)
6. General reaction: aA +bB cC + dD
G = [cGf(C) + dGf(D)] – [aGf(A) + bGf(B)]
Example 5
Calculate the standard free energy change for the reaction of calcium carbide with water. Might this reaction be used for the synthesis of acetylene (C2H2)?
CaC2(s) + 2 H2O(l) C2H2(g) + Ca(OH)2(s)
Gf (CaC2) = -64.8 kJ/mol
Gf (H2O(l)) = -237.2 kJ/mol
Gf (C2H2) = 209.2 kJ/mol
Gf (Ca(OH)2) = -898.6 kJ/mol
Free Energy Changes and Composition of the Reaction Mixture
Standard state conditions are unrealistic, the reaction itself will change the temperature and pressure, so what can we use to calculate the free energy change under non-standard state conditions?
G = G + RT ln Q
R = gas constant
T = temperature in Kelvins
Q = reaction quotient (Qc or Qp)
Example 6
Calculate the Free energy change for the formation of ethylene (C2H4) from carbon and hydrogen at 25C
when the partial pressures are 100 atm H2 and 0.10 atm
C2H4
2 C(s) + 2 H2(g) C2H4(g) G = 68.1 kJ/mol
Is the reaction spontaneous in the forward or reverse direction?
Free Energy and Chemical Equilibrium 1. When the RM is mostly reactants
Q<<1 RT lnQ <<0 G<0
2. When the RM is mostly products
Q>>1 RT lnQ >>0 G>0
3. G= -RT ln K
K = equilibrium constant Kc or Kp